Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{8n^2 - 40n}{n^2 - 14n + 40} \div \dfrac{7n - 35}{-8n + 80} $
Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{8n^2 - 40n}{n^2 - 14n + 40} \times \dfrac{-8n + 80}{7n - 35} $ First factor the quadratic. $p = \dfrac{8n^2 - 40n}{(n - 10)(n - 4)} \times \dfrac{-8n + 80}{7n - 35} $ Then factor out any other terms. $p = \dfrac{8n(n - 5)}{(n - 10)(n - 4)} \times \dfrac{-8(n - 10)}{7(n - 5)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ 8n(n - 5) \times -8(n - 10) } { (n - 10)(n - 4) \times 7(n - 5) } $ $p = \dfrac{ -64n(n - 5)(n - 10)}{ 7(n - 10)(n - 4)(n - 5)} $ Notice that $(n - 5)$ and $(n - 10)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -64n(n - 5)\cancel{(n - 10)}}{ 7\cancel{(n - 10)}(n - 4)(n - 5)} $ We are dividing by $n - 10$ , so $n - 10 \neq 0$ Therefore, $n \neq 10$ $p = \dfrac{ -64n\cancel{(n - 5)}\cancel{(n - 10)}}{ 7\cancel{(n - 10)}(n - 4)\cancel{(n - 5)}} $ We are dividing by $n - 5$ , so $n - 5 \neq 0$ Therefore, $n \neq 5$ $p = \dfrac{-64n}{7(n - 4)} ; \space n \neq 10 ; \space n \neq 5 $